Define $f(x, y, z) = \sin(xz)$. Let $\vec{a} = (5, 3, \pi)$ and $\vec{v} = \left( 0, 0, 1 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
When a directional derivative has a direction that equals $(1, 0, 0)$, $(0, 1, 0)$, or $(0, 0, 1)$, it becomes a regular partial derivative. Because $v = (0, 0, 1)$, the limit we want to find is the definition of $\dfrac{\partial f}{\partial z}$ evaluated at $(5, 3, \pi)$. Therefore: $ \lim_{h \to 0} \dfrac{f \left( 5, 3, \pi + h \right) - f(5, 3, \pi)}{h} = \dfrac{\partial f}{\partial z}(5, 3, \pi)$ $\begin{aligned} &\dfrac{\partial f}{\partial z} = x\cos(xz) \\ \\ &\dfrac{\partial f}{\partial z}(5, 3, \pi) = 5 \cos(5\pi) = -5 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -5$.